Solve 8x^2y^4-14x^2y^3+6xydiv2xy

$$8 { x }^{ 2 } { y }^{ 4 } -14 { x }^{ 2 } { y }^{ 3 } +6xy \div 2xy$$

Factor

$$\left(2y-3\right)\left(4y-1\right)x^{2}y^{2}$$

Solution Steps

Factor out xy.

$$xy\left(8xy^{3}-14xy^{2}+3xy\right)$$

Consider 8xy^{3}-14xy^{2}+3xy. Factor out xy.

$$xy\left(8y^{2}-14y+3\right)$$

Consider 8y^{2}-14y+3. Factor the expression by grouping. First, the expression needs to be rewritten as 8y^{2}+ay+by+3. To find a and b, set up a system to be solved.

$$a+b=-14 ab=8\times 3=24$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -14.

$$a=-12 b=-2$$

Rewrite 8y^{2}-14y+3 as \left(8y^{2}-12y\right)+\left(-2y+3\right).

$$\left(8y^{2}-12y\right)+\left(-2y+3\right)$$

Factor out 4y in the first and -1 in the second group.

$$4y\left(2y-3\right)-\left(2y-3\right)$$

Factor out common term 2y-3 by using distributive property.

$$\left(2y-3\right)\left(4y-1\right)$$

Rewrite the complete factored expression.

$$xyxy\left(2y-3\right)\left(4y-1\right)$$

Evaluate

$$\left(2y-3\right)\left(4y-1\right)\left(xy\right)^{2}$$